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Math Help - Backwards intergration

  1. #1
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    Backwards intergration

    \int\!\frac{1}{(1+x^2)^2}dx where x = tan u and u = arctan x

    I can't see where acrtan has come from. What part of the integrand is arctan?
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  2. #2
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    Re: Backwards intergration

    They chose x= tan(u) in order to use the trig identity 1+ tan^2(u)= sec^2(u). There is NO part of the integral that is arctan. You need u= arctan x when you want to convert the completed integral back to x.

    The point is that if you let x= tan(u) then 1+ x^2= 1+ tan^2 u= sec^2(u) while dx= sec^2(u)du so your integral becomes
    \int \frac{sec^2(u)du}{(sec^2(u))^2}= \int\frac{sec^2(u)du}{sec^4(u)}=\int \frac{du}{sec^2(u)}= \int cos^2(u)du. You can integrate that with the trig identity
    cos^2(u)= (1/2)(1+ cos(2u)).
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  3. #3
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    Re: Backwards intergration

    Excellent, thanks for the help.
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